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Old 04-25-2009, 09:33 PM
SooMe SooMe is offline
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It says "The prior distribution for sigma squared is expontential with mean 1.25." I take this to imply that sigma squared is therefore (1.25)^2 (variance = mean squared for exponential), but the solution says v = E(sigma^2) = 1.25. What... the F.
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Old 04-25-2009, 10:15 PM
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Gandalf Gandalf is offline
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Quote:
Originally Posted by SooMe View Post
It says "The prior distribution for sigma squared is expontential with mean 1.25." I take this to imply that sigma squared is therefore (1.25)^2 (variance = mean squared for exponential), but the solution says v = E(sigma^2) = 1.25. What... the F.
I'm not sure what the notation means, but as to what you wrote, I don't understand . sigma squared represents a variance, but the expected value of that variance is 1.25. That's what they tell you.

sigma squared isn't a single number; it has a distribution (an exponential distribution in particular). So since you cannot know what sigma squared is (its numeric value), you must decide what you do need to know about sigma squared: its expected value or its variance. The SOA solution says you want to know its expected value.

Last edited by Gandalf; 04-25-2009 at 10:20 PM..
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Old 04-26-2009, 11:26 AM
SooMe SooMe is offline
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I was hung up on the fact that it seemed as though they were saying the mean of the distribution was 1.25, not that the mean of the variance was 1.25. Alas.
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Old 05-18-2019, 12:59 AM
joelcheung joelcheung is offline
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I have a question on the part on the Variance of lambda. I used the formula for the discrete uniform distribution, because the question stated the range using the definite bracket, i.e. [ . ]. But based on the solution, the uniform distribution is continuous.

I believe the core of this confusion is the fact that the definite brackets are used.
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